Tukukkk

My goal is to conditionally index a data frame and change the values in a column for these indexes.

I intend on looking through the column 'A' to find entries = 'a' and update their column 'B' with the word 'okay.

group = ['a']



df = pd.DataFrame({"A": [a,b,a,a,c], "B": [NaN,NaN,NaN,NaN,NaN]})

>>>df

   A    B

0  a  NaN

1  b  NaN

2  a  NaN

3  a  NaN

4  c  NaN



df[df['A'].apply(lambda x: x in group)]['B'].fillna('okay', inplace=True)

This gives me the following error:

SettingWithCopyWarning:

A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
self._update_inplace(new_data)

Following the documentation (what I understood of it) I tried the following instead:

df[df['A'].apply(lambda x: x in group)].loc[:,'B'].fillna('okay', inplace=True)

I can't figure out why the reassignment of 'NaN' to 'okay' is not occurring inplace and how this can be rectified?

Thank you.

Try this with lambda:

Solution First:

>>> df

   A   B

0  a NaN

1  b NaN

2  a NaN

3  a NaN

4  c NaN

Using lambda + map or apply..

>>> df["B"] = df["A"].map(lambda x: "okay" if "a" in x else "NaN")

OR# df["B"] = df["A"].map(lambda x: "okay" if "a" in x else np.nan)

OR# df['B'] = df['A'].apply(lambda x: 'okay' if x == 'a' else np.nan)

>>> df

   A     B

0  a  okay

1  b   NaN

2  a  okay

3  a  okay

4  c   NaN

Solution second:

>>> df

   A   B

0  a NaN

1  b NaN

2  a NaN

3  a NaN

4  c NaN

another fancy way to Create Dictionary frame and apply it using map function across the column:

>>> frame = {'a': "okay"}

>>> df['B'] = df['A'].map(frame)

>>> df

   A     B

0  a  okay

1  b   NaN

2  a  okay

3  a  okay

4  c   NaN

Solution Third:

This is already been posted by @d_kennetz but Just want to club together, wher you can also do the assignment to both columns (A & B)in one shot:..

>>> df.loc[df.A == 'a', 'B'] = "okay"

If I understand this correctly, you simply want to replace the value for a column on those rows matching a given condition (i.e. where A column belongs to a certain group, here with a single value 'a'). The following should do the trick:

import pandas as pd



group = ['a']



df = pd.DataFrame({"A": ['a','b','a','a','c'], "B": [None,None,None,None,None]})

print(df)

df.loc[df['A'].isin(group),'B'] = 'okay'

print(df)

What we're doing here is we're using the .loc filter, which just returns a view on the existing dataframe.

First argument (df['A'].isin(group)) filters on those rows matching a given criterion. Notice you can use the equality operator (==) but not the in operator and therefore have to use .isin() instead).

Second argument selects only the 'B' column.
Then you just assign the desired value (which is a constant).

Here's the output:

   A     B

0  a  None

1  b  None

2  a  None

3  a  None

4  c  None

   A     B

0  a  okay

1  b  None

2  a  okay

3  a  okay

4  c  None

If you wanted to fancier stuff, you might want do the following:

import pandas as pd



group = ['a', 'b']



df = pd.DataFrame({"A": ['a','b','a','a','c'], "B": [None,None,None,None,None]})

df.loc[df['A'].isin(group),'B'] = "okay, it was " + df['A']+df['A']

print(df)

Which gives you:

   A                B

0  a  okay, it was aa

1  b  okay, it was bb

2  a  okay, it was aa

3  a  okay, it was aa

4  c             None